Question: Solve for $x$ : $5x^2 + 20x + 20 = 0$
Solution: Dividing both sides by $5$ gives: $ x^2 + {4}x + {4} = 0 $ The coefficient on the $x$ term is $4$ and the constant term is $4$ , so we need to find two numbers that add up to $4$ and multiply to $4$ The number $2$ used twice satisfies both conditions: $ {2} + {2} = {4} $ $ {2} \times {2} = {4} $ So $(x + {2})^2 = 0$ $x + 2 = 0$ Thus, $x = -2$ is the solution.